Let T P_2 rarr RR^3 be a transformation given by T(f(x))=[(f(0)),(f(1)),(2f(1))] (a)Then show that T is a linear transformation. (b)Find and describe the kernel(null space) of T i.e Ker(T) and range of T. (c)Show that T is one-to-one.

Maiclubk 2021-03-02 Answered
Let T \(\displaystyle{P}_{{2}}\rightarrow\mathbb{R}^{{3}}\) be a transformation given by
\(\displaystyle{T}{\left({f{{\left({x}\right)}}}\right)}={\left[\begin{array}{c} {f{{\left({0}\right)}}}\\{f{{\left({1}\right)}}}\\{2}{f{{\left({1}\right)}}}\end{array}\right]}\)
(a)Then show that T is a linear transformation.
(b)Find and describe the kernel(null space) of T i.e Ker(T) and range of T.
(c)Show that T is one-to-one.

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Expert Answer

Elberte
Answered 2021-03-03 Author has 17940 answers

The map T : \(\displaystyle{P}_{{2}}\rightarrow\mathbb{R}^{{3}}\) be a transformation given by
\(\displaystyle{T}{\left({f{{\left({x}\right)}}}\right)}={\left[\begin{array}{c} {f{{\left({0}\right)}}}\\{f{{\left({1}\right)}}}\\{2}{f{{\left({1}\right)}}}\end{array}\right]}\).
(a)Let f,\(\displaystyle{g}\in{P}_{{2}}\) and k is a real number. Then
\(\displaystyle{T}{\left({f{{\left({x}\right)}}}+{k}{g{{\left({x}\right)}}}\right)}={T}{\left({f{{\left({f}+{k}{g}\right)}}}{\left({x}\right)}\right)}\)
\(\displaystyle={\left[\begin{array}{c} {\left({f}+{k}{g}\right)}{\left({0}\right)}\\{\left({f}+{k}{g}\right)}{\left({1}\right)}\\{2}{\left({f}+{k}{g}\right)}{f{{\left({1}\right)}}}\end{array}\right]}\)
\(\displaystyle={\left[\begin{array}{c} {f{{\left({0}\right)}}}+{k}{g{{\left({0}\right)}}}\\{\left({f{{\left({1}\right)}}}+{k}{g{{\left({1}\right)}}}\right)}\\{2}{f{{\left({1}\right)}}}+{2}{k}{g{{\left({1}\right)}}}\end{array}\right]}\)
\(\displaystyle={\left[\begin{array}{c} {f{{\left({0}\right)}}}\\{f{{\left({1}\right)}}}\\{2}{f{{\left({1}\right)}}}\end{array}\right]}+{\left[\begin{array}{c} {k}{g{{\left({0}\right)}}}\\{k}{g{{\left({1}\right)}}}\\{2}{k}{g{{\left({1}\right)}}}\end{array}\right]}\)
\(=T(f(x))+kT(g(x)).\)
Therefore T:\(\displaystyle{P}_{{2}}\rightarrow\mathbb{R}^{{3}}\) is a linear transformation.
(b)
\(\displaystyle{K}{e}{r}{\left({T}\right)}={\left\lbrace{f}\in{P}_{{2}}:{T}{\left({f{{\left({x}\right)}}}\right)}={\left[\begin{array}{c} {0}\\{0}\\{0}\end{array}\right]}\right\rbrace}\)
\(\displaystyle={\left\lbrace{f}\in{P}_{{2}}:{\left[\begin{array}{c} {f{{\left({0}\right)}}}\\{f{{\left({1}\right)}}}\\{2}{f{{\left({1}\right)}}}\end{array}\right]}={\left[\begin{array}{c} {0}\\{0}\\{0}\end{array}\right]}\right\rbrace}\)
\(\displaystyle={\left\lbrace{f}\in{P}_{{2}}:{f{{\left({0}\right)}}}={0},{f{{\left({1}\right)}}}={0},{2}{f{{\left({1}\right)}}}={0}\right\rbrace}\)
\(\displaystyle={\left\lbrace{f}\in{P}_{{2}}:{f{{\left({0}\right)}}}={0},{f{{\left({1}\right)}}}={0}\right\rbrace}\)
\(\displaystyle={\left\lbrace{f{{\left({x}\right)}}}={a}{x}{\left({x}-{1}\right)},{a}\in\mathbb{R}\right\rbrace}\)
Therefore Nullity \((T)=1.\)
Again Range(T) \(\displaystyle={\left\lbrace{T}{\left({f{{\left({x}\right)}}}\right)}:{f}\in{P}_{{2}}\right\rbrace}={s}{p}{a}{n}{\left\lbrace{f{{\left({0}\right)}}},{f{{\left({1}\right)}}}\right\rbrace}\). Therefore
Rank \((T)=2\).
(c)Since Nullity \((T)=1\) therefore T is not one-one.

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